Limit of difference of integral and sum $f:[0,1]\rightarrow\mathbb R$ and $f\in C^1$, then the limit $\lim_{n\rightarrow\infty} n(\int_{0}^{1}f(x)dx-\frac{1}{n}\sum_{k=1}^{n}f(\frac{k-1}{n}))$ exists.
I guess the kernel lies in the sum because then I can write the sum as an integral but I do not know how. 
 A: Every continuous function is integrable an hence all Riemann sums converge to the value of the integral, notice that the sum is just a Riemann sum of a partition of [0,1]. Now you have an estimation of the error which is made by the Riemann sum. Here use that the derivative is continuous too and hence takes a maximum in [0,1] (check your definition of differentiable on a closed interval, if you have defined it differently this might not work).
When you look at the intervalls $I=\left[\frac{i}{n},\frac{i+1}{n}\right]$ you have
\[ \left|\int_{\frac{i}{n}}^\frac{i+1}{n} f(x) \; \mathrm{d}x -\frac{1}{n} f\left(\frac{i}{n}\right)\right| \leq \frac{1}{n^2} \cdot \sup_{\xi \in I} f'(\xi)\]
A little more explanation, with the fundamental theorem of calculus we have
\[\int_{\frac{i}{n}}^\frac{i+1}{n} f(x)\; \mathrm{d}x= \frac{1}{n} \cdot f(\xi_1)\]
with $\xi_1 \in \left(\frac{i}{n}, \frac{i+1}{n}\right)$. 
So we know that 
\[\int_{\frac{i}{n}}^\frac{i+1}{n} f(x) \; \mathrm{d}x -\frac{1}{n} f\left(\frac{i}{n}\right) = \frac{1}{n} \left(f(\xi_1)-f(\tfrac{i}{n}) \right) \]
With mean value theorem you get 
\[ \frac{1}{n} \left(f(\xi_1)-f(\tfrac{i}{n}) \right)==\frac{1}{n} \cdot \left(\xi_1-\frac{i}{n} \right) \cdot f'(\xi_2)\leq \frac{1}{n^2} f'(\xi_2)\]
with $\xi_2 \in \left( \frac{i}{n} , \xi_i\right)$
Using all this, we have 
\[ \lim_{n\to \infty} n \cdot \left( \int_0^1 f(x)\; \mathrm{d}x - \frac{1}{n} \sum_{k=1}^n f(\tfrac{i-1}{n})\right) = \lim_{n\to \infty} n \cdot \left( \sum_{i=1}^n \frac{1}{n} \cdot \left( \xi_i-\frac{i}{n}\right) f'(\xi_i) \right)\]
Now we look at the absolute values and have 
\begin{align*}
 \lim_{n\to \infty} n \cdot \left( \sum_{i=1}^n \frac{1}{n} \cdot \left( \xi_i-\frac{i}{n}\right) f'(\xi_i) \right)& \leq \lim_{n\to \infty} \sum_{i=1}^n  \left( \xi_i -\frac{1}{n} \right) |f'(\xi_i)|\\
& \leq \lim_{n\to \infty} \sum_{i=1}^n \frac{1}{n} |f'(\xi_i)| \\
&\leq \sup_{\xi \in [0,1]} |f'(\xi)| \end{align*}
